Field $\mathbb{Q}(\alpha)$ with $\alpha=\sqrt[3]7+2i$

For $2:$

Given that: $$x^6+12x^4-14x^3+48x^2+168x+113=(x^3-12x-7)^2+(6x^2-8)^2,$$ you have that:

$$\left(\frac{\alpha^3-12\alpha-7}{6\alpha^2-8}\right)^2=-1,$$

at least as long as you can show that $6\alpha^2\neq 8.$

This means that $$\frac{\alpha^3-12\alpha-7}{6\alpha^2-8}=\pm i$$ and hence $i\in\mathbb Q(\alpha)$.

For (3-4): Since $\mathbb Q(i)\subseteq \mathbb Q(\alpha)$ and $\mathbb Q(\sqrt[3]7)\subseteq \mathbb Q(\alpha)$, you have that $[Q(\alpha):\mathbb Q]$ must be divisible to $3$ and $2$, and thus be divisible by $6$.

But your polynomial also gives us that $[\mathbb Q(\alpha):\mathbb Q]\leq 6$.

So $\mathbb Q(\alpha):\mathbb Q]=6$, and thus the minimal polynomial of $\alpha$ must be degree $6$.

Your proof of the irreducibility of $f(x)=x^6 +12x^4-14x^3 +48x^2+168x+113$ is not correct. Here’s a way of doing it, though. Over $\Bbb Q(i)$, we have the factorization $$ f(x)=(x^3-6ix^2-12x+8i-7)(x^3+6ix^2-12x-8i-7)\,, $$ which can be found by various methods. These cubic polynomials are $\Bbb Q(i)$-irreducible, since their roots are not in $\Bbb Q(i)$, so this is the $\Bbb Q(i)$-factorization into irreducibles, unique.

Now assume $f$ had a nontrivial $\Bbb Q$ factorization. This would be a $\Bbb Q(i)$-factorization, and hence the product of two quadratics in the displayed formula. But these are not $\Bbb Q$-polynomials, contradicting our assumption. So $f$ is irreducible.